Let $l/K$ be a finite(degree $d$) ramified Galois extension of local fields and $E/K$ be an elliptic curve.Let $l$ be a residue field of $L$.
Let $E_1$ be a reduction of kernel. Why does there exists an exact sequence $E_1(K)/\text{trace}(E_1(L))\to E(K)/\text{trace}(E(L))\to \tilde{E}(l)/d\tilde{E}(l)\to 0$?
I think the left hand side should be a kernel of the middle map, but I cannot grasp the middle surjective map. Why is the kernel of the middle map is $E_1(L)/\text{trace}(E_1(K))$?
The middle map is the reduction map mod the maximal ideal. It's well-defined, because the extension is totally ramified, so an element of the form $\text{tr}(x) = \sum_{x \in Gal(L/K)} x^\sigma$ maps to a multiple of $d$ since the local extension is trivial.
An element is in the kernel of this map iff its reduction is a multiple of $d$. It's clear that anything whose reduction is $0$ is in this kernel, hence the left arrow, and we need to show that the image of this arrow is the whole kernel of the next one.
If $P$ is a point in $E(K)$ representing a class in the kernel of this map, then the reduction of $P$ is a multiple of $d$, say $\bar{P} = d\bar{Q}$.
Lifting $\bar{Q}$ to some choice of $Q$ (using the surjectivity of the reduction map!) gives that $P - dQ$ is in the kernel of the reduction map. But $P - dQ$ is equivalent to $P$ up to a trace of something from $E(L)$, since in particular $Q \in E(L)$ and $\text{trace}(Q) = dQ$.